Name
ID
MATH 251
Final Exam
Sections 515
Solutions
1-9
/45
Fall 2012
10
/15
P. Yasskin
11
/25
12
/15
Total
/100
Multiple Choice: (5 points each. No part credit.)
1. Points A, B, C and D are the vertices of a parallelogram traversed in order.
If A
2, 4, 1 , B
a.
2, 3, 3
b.
4, 1, 3
c.
0, 9, 1
d.
6, 5, 1
e.
4, 9 , 0
2
SOLUTION:
C B AD
3, 2, 0
and D
1, 3, 2 , then C
Correct Choice
AD D
3, 2, 0
A
1, 3, 2
2, 4, 1
1, 1, 3
2, 3, 3
2. Which vector is perpendicular to the surface x 2 z 3
a.
12, 24, 43
b.
6, 12, 43
c.
6, 12, 43
d.
6, 12, 11
e.
12, 24, 22
SOLUTION:
12, 24, 22
f
1, 1, 3
y3z2
1 at the point
3, 2, 1 ?
Correct Choice
x2z3
y3z2
f
2xz 3 , 3y 2 z 2 , 3x 2 z 2
2y 3 z
f
3, 2,1
6, 12, 11
2 6, 12, 11
1
3. Find the point on the elliptic paraboloid R t,
2 3
, 2, 3
5
5 5
N
c.
3 , 3 ,2
2
3 3 , 1, 2
2
3, 2 3 , 5
d.
3 3 , 2, 5
a.
b.
where a unit normal is
.
Correct Choice
3, 2 3 , 5
e.
k
î
SOLUTION:
16t 4
N
4t 2
R 2,
t2
3t cos , 2t sin , 1
9
et
3 cos ,
2 sin ,
2t
e
3t sin , 2t cos ,
0
36t 2
5
2t 4t 2
t
2
3 2 cos
6
9
6
, 2 2 sin
2t sin ,
4t 2 9
3
cos
2
3
6
,4 1 ,5
2
2
2 3
6
4t 2 cos , 4t 2 sin , 6t
2t cos ,
4t 2 9
N
4 cos
N
22
,1
2 3
, 2, 3
5 5
5
3
4t
2
9
6
3 3 , 2, 5
9 y2
3
1
dx dy
1 x2 y2
HINT: Plot the region of integration and convert to polar coordinates.
4. Compute
9 y2
0
ln 10
2
b. ln 10
a.
Correct Choice
arctan 3
2
d. arctan 3
c.
e.
arctan 10
SOLUTION:
0
9
y
y2
3
x
9
y
y2
2
or
0
3
r
3
1
0
-3
3
0
9 y2
9 y2
1
1
x2
3
y2
dx dy
0
0
r
1
r2
dr d
2
ln 1
r2
-2
3
0
2
-1
0
1
2
3
x
ln 10
2
e t , 2 t, e
5. Find the mass of a wire in the shape of the curve r t
is
t
for
1
t
1 if the density
x.
2
a. e
e 2
2
e 2
2
e 2
2
2
2
b. e
2
2
c. e
2
d. e 2
e
2
e. e 2
e
2
2
Correct Choice
v
et, 2 , e
2
SOLUTION:
1
M
1
1
e2
2
et et
x |v | dt
ds
1
t
e 2t
|v |
e
t
1
1
e 2t
dt
1
e 2
2
2
e
2t
et
e 2t
2
1 dt
1
e2
2
6. Find the plane tangent to graph of z
e 2
2
e
t
1
t
1
2
x cos y
sin y at
2,
.
What is the z-intercept?
a.
4
b. 4
c.
4
d. 4
e.
Correct Choice
SOLUTION:
f
f x x, y
cos y
f y x, y
x sin y
z f 2,
f x 2,
tan plane:
z
z-intercept:
x
x, y
x cos y sin y
f 2,
2 cos
sin
f x 2,
cos
1
cos y
f y 2,
2 sin
cos
1
x 2 f y 2,
y
2 1x 2 1y
y 0
z
2 1 2 1
2
3
t, t 2 , t 3
7. Find the plane perpendicular to the curve r t
at the point
1, 1, 1 .
What is the z-intercept?
a. 5
b. 4
c. 3
Correct Choice
d. 2
e. 1
1, 1, 1
at
t 1
SOLUTION:
P rt
2
vt
1, 2t, 3t
N v1
1, 2, 3
Plane:
N X N P
x 2y 3z 1 1 2 1
z-intercept:
x y 0
3z 6
z 2
8. Compute
y 2 , 4xy
F ds for F
3 1
6
from
2, 4 to 2, 4 followed by the line segment from
HINT: Use Green’s Theorem.
a. 256
c.
d.
e.
SOLUTION:
F ds
2
2
xQ
x2
2
16
x 4 dx
yP
16x
2
9. Compute
Q dy with P
P dx
4
F ds
2
back to
2, 4
2, 4 .
Correct Choice
5
768
5
64
3
128
3
0
b.
x2
along the piece of the parabola y
F ds for F
4
dy dx
2
x2
2
x5
5
y 2 and Q
2 32
2
4xy 2 , 4x 2 y
2
4y
4xy. By Green’s Theorem,
y2
2y dy dx
2
32
5
4
x2
dx
256
5
along the line segment from
to
1, 2
3, 1 .
HINT: Find a scalar potential.
a. 4
b. 10
Correct Choice
c. 20
d. 24
e. 26
SOLUTION:
By the F.T.C.C.
F
4xy 2 , 4x 2 y
f for f
2x 2 y 2 since
3,1
F ds
f ds
f 3, 1
f 1, 2
23
4xy 2 and
xf
2
1
2
21
2
2
yf
2
4x 2 y
10
1,2
4
Work Out: (Points indicated. Part credit possible. Show all work.)
10. (15 points) A 166 cm piece of wire is cut into 3 pieces of lengths a, b and c.
a.
4
The piece of length b is folded into a rectangle of length L 1
The piece of length a is folded into a square of side s
b and width W 1
3
3c and width W 2
8
The piece of length c is folded into a rectangle of length L 2
b.
6
c.
8
Find a, b and c so that the total area is a minimum.
What is the total area?
Minimize the total area:
subject to the constraint:
A
g
s2
a
L1W1
L2W2
b
166
c
METHOD 1: Lagrange Multipliers:
a , b , 3c
A
g
1, 1, 1
8 9 32
a
b
3c
A
g
8
9
32
3c
27c
a
b
4
32
3c
27c c
Constraint:
166 a b c
4
32
3c
27c
a
48
b
54
4
32
48 2
54 2
3 64 2
A
498
16
18
64
a2
16
b2
18
3c 2
64
a
8
83 c
32
b
9
c
3c
32
64
METHOD 2: Eliminate a Variable:
c
166 a b
3 166 a b 2
a2
b2
A
16
18
64
3
166
a
b
a
7 a
3 b 249
Aa
8
32
32
32
16
3
166
a
b
b
3 a
59 b 249
Ab
32
9
32
288
16
59
3 1 7 2 : 9b 7
b
4 498
9
59 1 3 2 : 59 7a 9a
59 3 498
9
9
9
c 166 a b 166 48 54 64
48 2
54 2
3 64 2
A
498
16
18
64
0
0
332 b
9
7a
3a
3b
59 b
9
1992
332 a
5312
9
3
(1)
498
(2)
498
b
54
a
48
5
11.
(25 points) Verify Gauss’ Theorem
F dV
V
V
for the vector field F
hemisphere 0
xz, yz, x
z
x2
4
2
y
2
F dS
and the solid
y2 .
Be careful with orientations. Use the following steps:
First the Left Hand Side:
a. Compute the divergence:
F
z
z
0
2z
b. Express the divergence and the volume element in the appropriate coordinate system:
F
2 cos
2
dV
sin d d d
c. Compute the left hand side:
2
/2
2
F dV
0
V
0
/2
sin 2
2
2
2
2 cos
0
4
2
0
/2
sin d d d
2
0
2
1
2
2
0
8
2
cos sin d
2
3
d
0
8
Second the Right Hand Side:
The boundary surface consists of a hemisphere H and a disk D with appropriate
orientations.
d. Parametrize the disk D:
R r,
r cos , r sin , 0
e. Compute the tangent vectors:
cos ,
er
e
r sin ,
sin ,
0
r cos ,
0
f. Compute the normal vector:
N
î0
r sin 2
k r cos 2
0
0, 0, r
This is up. Need down. Reverse: N
g. Evaluate F
F
R r,
xz, yz, x 2
y2
0, 0, r
on the disk:
0, 0, r 2
h. Compute the dot product:
F N
r3
i. Compute the flux through D:
2
2
F dS
D
0
0
r 3 dr d
2
r4
4
2
8
0
6
j. Parametrize the hemisphere H:
R
,
2 sin cos , 2 sin sin , 2 cos
k. Compute the tangent vectors:
e
2 cos cos ,
2 cos sin ,
2 sin
e
2 sin sin ,
2 sin cos ,
0
l. Compute the normal vector:
N
î 4 sin 2 cos
4 sin 2 sin
k 4 sin cos sin 2
4 sin cos cos 2
4 sin 2 cos , 4 sin 2 sin , 4 sin cos
This is oriented correctly as up and out.
m. Evaluate F
F
xz, yz, x 2
y2
on the hemisphere:
4 sin cos cos , 4 sin cos sin , 4 sin 2
R ,
n. Compute the dot product:
16 sin 3 cos cos 2
F N
16 sin 3 cos
16 sin 3 cos sin 2
16 sin 3 cos
16 sin 3 cos
32 sin 3 cos
o. Compute the flux through H:
2
/2
F dS
0
C
32 sin 3 cos d d
2 8 sin 4
0
/2
0
16
p. Compute the TOTAL right hand side:
F dS
V
F dS
D
F dS
8
16
8
which agrees with (c).
H
7
12.
(15 points) Compute
F dS for F
y, x, z
over the "clam shell" surface parametrized by
R r,
for r
r cos , r sin , r sin 6
2 oriented upward.
HINTS: Use Stokes Theorem.
What is the value of r on the boundary?
SOLUTION:
Boundary: r
Fr
F v
2
r
2 cos , 2 sin , 2 sin 6
v
2 sin , 2 cos , 12 cos 6
2 sin , 2 cos , 2 sin 6
4 sin 2
4 cos 2
24 sin 6
cos 6
2
F dS
F ds
4
F vd
0
24 sin 6
cos 6
2
4
0
24 sin 6
cos 6
d
4
2 sin 2 6
2
8
0
8